Computer Data Communication

1. Computer Data Communication Extra Slide I

2. Contents • Frequency and Period • Bit Rate • Throughput • Fourier Analysis • Latency (Delay) • Data Rate and Signal Rate

3. Frequency and Period • Period refers to the amount of time in seconds a signal needs to complete 1 cycle. • Frequency refers to the number of periods in 1 s. • Period is the inverse of frequency and frequency is the inverse of period. f = 1/T

4. Frequency and Period • Period is formally expressed in seconds. • Frequency is formally expressed in Hertz (Hz), which is cycle per second • If a signal does not change at all, its frequency is zero. • If a signal changes instantaneously, its frequency is infinite.

5. Frequency and Period (Example) • The period of a signal is 100 ms.What is its frequency in kilohertz? 100 ms = 100 x 10-3 s = 10-1 s f = 1/T = 1/10-1 = 10 Hz = 10 x 10-3 kHz = 10-2 kHz

6. Bit Rate • The bit rate is the number of bits sent in 1s, expressed in bits per second (bps)

7. Bit Rate (Example) • Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel? • Note:A page is an average of 24 lines with 80 character in each line. One character requires average 8 bits. Bit rate = 100 x 24 x 80 x 8 = 1,636,00 bps = 1.636 Mbps

8. Bit Rate (Example) • A digitized voice channel is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). Assume that each sample requires 8 bits. What is the required bit rate? Bit rate = 2 x 4000 x 8 = 64,000 bps = 64 kbps

9. Bit Rate (Example) • HDTV uses digital signal to broadcast high quality video signals. The HDTV Screen is normally a ratio of 16:9 (in contrast to 4:3 for regular TV), which means the screen is wider. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty four bits represent one color pixel. Bit rate = 1920 x 1080 x 30 x 24 = 1,492,992,000 bps = 1.5 Gbps

10. Throughput • The throughput is a measure of how fast we can actually send data through a network. • A link may have a bandwidth of B bps, but we can only send T bps through this link with T always less than B. • The bandwidth is a potential measurement of a link; the throughput is an actual measurement of how fast we can send data. • E.g. • We may have a link with a bandwidth of 1 Mbps, but the devices connected to the end of the link may handle only 200 kbps. This means that we cannot send more than 200 kbps through this link. • A highway designed to transmit 1000 cars per minute from one point to another. However, if there is a congestion on the road, this figure may be reduced to 100 cars per minute. The bandwidth is 1000 cars per minute; the throughput is 100 cars per minute.

11. Throughput (Example) • A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minutes with each frame carrying an average of 10,000 bits. What is the throughput of this network? Throughput = (12,000 x 10,000) / 60 = 2 Mbps

12. Fourier Analysis • Based on Fourier Analysis, a digital signal is a composite analog signal. • Fourier analysis can be used to decompose a digital signal. • If the digital signal periodic, the decomposed signal has a frequency domain representation with an infinite bandwidth and discrete frequencies. • If the digital signal is nonperiodic, the decomposed signal still has an infinite bandwidth, but the frequencies are continuous.

13. Latency (Delay) • The latency or delay defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source. • Latency is made of four components: propagation time, transmission time, queuing time and processing delay.

14. Propagation Time • Propagation time measures the time required for a bit to travel from the source to the destination. • The propagation time is calculated by dividing the distance by the propagation speed. Propagation time = Distance _ Propagation speed

15. Propagation Time (Example) • What is the propagation time if the distance between the two points in 12,00 km? Assume the propagation speed to be 2.4 x 108 ms in cable. Propagation time = (12,000 x 1000) / 2.4 x 108 = 50 ms

16. Transmission Time • There is a time between the first bit leaving the sender and the last bit arriving at the receiver. • The time required for transmission of a message depends on the size of the message and the bandwidth of the channel. Transmission time = Message size Bandwidth

17. Transmission Time (Example) • A device is sending out data at the rate of 1000 kbps. How long does it take to send out 10 bits? Transmission Time = 10 bits / 106 bps = 0.00001 s

18. Queuing Time • Queuing time is the time needed for each intermediate or end device to hold the message before it can be processed. • The queuing time is not a fixed factor; it changes with the load imposed on the network. • An intermediate device, such as a router, queues the arrived messages and processes them one by one. If there are many messages, each message will have to wait.

19. Processing Delay • Processing delay also known as Bandwidth-Delay product Number of bits = bandwidth x delay E.g.: How many bits can fits on a link with a 2 ms delay if the bandwidth of the link is 10 Mbps? Number of bits = 10 Mbps x 2 ms = 10,000,000 bps x 0.002 s = 12000 bits

20. Data Rate vs. Signal Rate • The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps) • The signal rate is the number of signal elements sent in 1s. The unit is the baud. • One goal in data communication is to increase the data rate while decreasing the signal rate. • Increasing the data rate increases the speed of transmission; decreasing the signal rate decreases the bandwidth requirement.